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------------------------------------------------ How many mL of a 16 M NH3 solution are needed to prepare 2.00 L of a 2.00 M solution?

The molarity is given as 16 M, the volume of 16 M [tex]NH3[/tex] solution required to prepare 2.00 L of a 2.00 M solution is 0.25 liters or 250 ml.

To prepare a 2.00 M solution of [tex]NH3[/tex] (ammonia) with a volume of 2.00 L, we can use the formula for calculating the amount of solute needed:

moles of solute = Molarity × Volume (in liters)

Molarity of the desired solution (M1) = 2.00 M

Volume of the desired solution (V1) = 2.00 L

Molarity of the stock [tex]NH3[/tex] solution (M2) = 16 M

Let's calculate the moles of [tex]NH3[/tex] needed:

moles of [tex]NH3[/tex] = M1 × V1

moles of [tex]NH3[/tex] = 2.00 M × 2.00 L

moles of [tex]NH3[/tex] = 4.00 moles

Now, we need to calculate the volume of the 16 M [tex]NH3[/tex] solution containing 4.00 moles of [tex]NH3[/tex]:

Volume of 16 M [tex]NH3[/tex]solution (V2) = moles of [tex]NH3[/tex] ÷ M2

Volume of 16 M [tex]NH3[/tex] solution (V2) = 4.00 moles ÷ 16 M

Volume of 16 M [tex]NH3[/tex] solution (V2) = 0.25 L

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